# Showing that $\operatorname{rank}(A'A + B'B) =\operatorname{rank}(A)+\operatorname{rank}(B)\ \rm{given}\ AB'=0$

If $A$ is an $m\times n$ matrix and $B$ is a $s\times n$ matrix such that $AB'=0$, show that $$\operatorname{rank}(A'A + B'B) = \operatorname{rank}(A)+\operatorname{rank}(B).$$ I have tried using the result $\operatorname{rank}(P+Q)<\operatorname{rank}(P)+\operatorname{rank}(Q)$(P an Q are both $p\times q$) but it is not working.

$\matrix A'$ is A transpose.

-
Although not essential, please clarify the meaning of $X'$. (Transpose? Conjugate transpose?) Also, would you please explain the meaning of $A+B$ when $m\neq s$? –  user1551 Jul 12 '13 at 10:14
If $A$ and $B$ don't have the same dimensions, how do you defined $A+B$? –  xavierm02 Jul 12 '13 at 10:19
@xavierm02 Note that $A$ and $B$ have the same number of columns. And since (clearly? most likely?) $A'$ is for $A^*$ or $A^T$, $A^*A$ and $B^*B$ are both $n\times n$. Hmmm, but the question was edited... –  1015 Jul 12 '13 at 13:29
I assume $A'$ means $A^*$, or just $A^T$ in the real case. Then try to use $\ker A^*A=\ker A$ and $\mbox{im } A^*A\perp \mbox{im } B^*B$. –  1015 Jul 12 '13 at 13:35
Note also that $A^*A$ and $B^*B$ are positive semidefinite and commute. So you can diagonalize them simultaneously (in an orthonormal basis). That their images are orthogonal will help you then. –  1015 Jul 12 '13 at 13:50